Question

Prove that √sec²∅ + Cos²∅ = tan ∅ + cos ∅.​

Answer 1

$\huge\underline\mathrm{SOLUTION:-}$

$\mathsf { L.H.S = \sqrt{ \sec {}^{2} (\theta) + \cos \: es {}^{2} (\theta) } }$

$\mathsf { = \sqrt{ \frac{1}{ \cos {}^{2} (\theta) } + \frac{1}{ \sin {}^{2} (\theta) } } }$

$\mathsf {= \sqrt{ \frac{ \sin {}^{2} (\theta) + \cos {}^{2} (\theta) }{ \sin {}^{2} (\theta). \cos {}^{2} (\theta) } } }$

$\mathsf { = \sqrt{ \frac{1}{ \sin {}^{2} (\theta) . \cos {}^{2} (\theta) } } }$ [∵ Sin²∅ + Cos²∅ = 1]

$\mathsf {= \frac{1}{ \sin(\theta). \cos(\theta) } }$

$\mathsf { = \frac{ \sin {}^{2} (\theta) + \cos {}^{2} (\theta) }{ \sin(\theta). \cos(\theta) }}$ [∵ 1 = Sin²∅ + Cos²∅]

$\mathsf { = \frac{ \sin {}^{2} (\theta) }{ \sin(\theta). \cos(\theta) } + \frac{ \cos {}^{2} (\theta) }{ \sin(\theta). \cos(\theta) } }$

$\mathsf { = \frac{ \sin(\theta) }{ \cos(\theta) } + \frac{ \cos(\theta) }{ \sin(\theta) }}$

$\mathsf {= \tan(\theta) + \cot(\theta) = R.H.S}$

• Hence Proved.

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